Problem: Simplify and expand the following expression: $ \dfrac{2}{4y - 16}+ \dfrac{3}{4y + 24}- \dfrac{1}{y^2 + 2y - 24} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{2}{4y - 16} = \dfrac{2}{4(y - 4)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4y + 24} = \dfrac{3}{4(y + 6)}$ We can factor the quadratic in the third term: $ \dfrac{1}{y^2 + 2y - 24} = \dfrac{1}{(y - 4)(y + 6)}$ Now we have: $ \dfrac{2}{4(y - 4)}+ \dfrac{3}{4(y + 6)}- \dfrac{1}{(y - 4)(y + 6)} $ The least common multiple of the denominators is: $ 16(y - 4)(y + 6)$ In order to get the first term over $16(y - 4)(y + 6)$ , multiply by $\dfrac{4(y + 6)}{4(y + 6)}$ $ \dfrac{2}{4(y - 4)} \times \dfrac{4(y + 6)}{4(y + 6)} = \dfrac{8(y + 6)}{16(y - 4)(y + 6)} $ In order to get the second term over $16(y - 4)(y + 6)$ , multiply by $\dfrac{4(y - 4)}{4(y - 4)}$ $ \dfrac{3}{4(y + 6)} \times \dfrac{4(y - 4)}{4(y - 4)} = \dfrac{12(y - 4)}{16(y - 4)(y + 6)} $ In order to get the third term over $16(y - 4)(y + 6)$ , multiply by $\dfrac{16}{16}$ $ \dfrac{1}{(y - 4)(y + 6)} \times \dfrac{16}{16} = \dfrac{16}{16(y - 4)(y + 6)} $ Now we have: $ \dfrac{8(y + 6)}{16(y - 4)(y + 6)} + \dfrac{12(y - 4)}{16(y - 4)(y + 6)} - \dfrac{16}{16(y - 4)(y + 6)} $ $ = \dfrac{ 8(y + 6) + 12(y - 4) - 16} {16(y - 4)(y + 6)} $ Expand: $ = \dfrac{8y + 48 + 12y - 48 - 16}{16y^2 + 32y - 384} $ $ = \dfrac{20y - 16}{16y^2 + 32y - 384}$ Simplify: $ = \dfrac{5y - 4}{4y^2 + 8y - 96}$